Question

Explain why the following reaction occurs:
Zn + 2HCl → ZnCl2 + H,

Answer 1

The reaction
`\rm Zn\, (s) + 2\, HCl \, (aq) \to ZnCl_2\, (aq) + H_2\, (g)` is spontaneous (under standard conditions) because the cell potential is positive.

Step-by-step explanation:

In this reaction:

• Metallic zinc
  `\rm Zn\, (s)` is oxidized to produce zinc ions
  `\rm Zn^(2+)`.
• Hydrogen ions
  `\rm H^(+)` is reduced to produce hydrogen gas
  `\rm H_2\, (g)`.

The anode is where the oxidation half-reaction takes place. In this case, the anode half-reaction is
`\rm Zn\, (s) \to Zn^(2+) \, (aq) + 2\, e^(-)`.

The cathode is where the reduction half-reaction takes place. In this case, the cathode half-reaction is
`\rm 2\, H^(+)\, (aq) + 2\, e^(-) \to H_2\, (g)`.

Look up the standard reduction potential for these two half-reactions:

• 
  `E^(\circ)_{\text{reduction, anode}} = \rm -0.7618\; V` for the reaction
  `\rm Zn^(2+) \, (aq) + 2\, e^(-) \rightleftharpoons Zn\, (s)`.
• 
  `E^(\circ)_{\text{reduction, cathode}} = \rm 0\; V` for the reaction
  `\rm H^(+) \, (aq) + e^(-) \rightleftharpoons (1)/(2)H_2\, (g)`.

The standard cell potential is equal to the standard reduction potential at the cathode minus that at the anode:

`E^(\circ)_\text{cell} = E^(\circ)_{\text{reduction, cathode}} - E^(\circ)_{\text{reduction, anode}} = \rm 0.7618\; V`.

The electrochemical reaction in the cell is spontaneous if and only if the cell potential is positive. Therefore, the reaction
`\rm Zn\, (s) + 2\, HCl \, (aq) \to ZnCl_2\, (aq) + H_2\, (g)` is spontaneous under standard conditions.